BD is a perpendicular bisector of AC. Which statement can NOT always be proven?

∠ADB ≅ ∠CDB

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AD ≅ CD

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ΔABD ≅ ΔCBD

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BD ≅ DC

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Solution

The correct option is D
BD ≅ DC

$A D ≅ C D$ as $B D$ is a perpendicular bisector and using $S A S$ criteria we can prove that $Δ A B D ≅ Δ C B D$ .

$B D$ and $C D$ are not the corresponding sides and we do not have any other information about the sides.

Hence, it is not always possible to prove that $B D ≅ D C$ .

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Similar questions

BD is a perpendicular bisector of AC. Which statement can NOT always be proven?

Q. In triangle ABC, BD is the perpendicular bisector of BC meets AC in point D the BD is angle bisector of angle ABC if AD=9 DC=7 then find the area of triangle ABD.

$Δ A B D$ is a right triangle right-angled at A and AC \perp BD. Show that
(i) $A B^{2} = B C . B D$
(ii) $A C^{2} = B C . D C$
(iii) $A D^{2} = B D . C D$
(iv) $\frac{A B^{2}}{A C^{2}} = \frac{B D}{D C}$

Q. In Fig.,

A B D

is a triangle right angled at

A

and

A C

⊥

BD. Show that:
(i)

A B^{2} = B C . B D

(ii)

A C^{2} = B C . D C

(iii)

A D^{2} = B D . C D

In $△$ ABC, B is a right angle, and BD is perpendicular to AC and DE is perpendicular to BC. Then $A D . D C =$

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BD is a perpendicular bisector of AC. Which statement can NOT always be proven?

The correct option is D BD ≅ DC AD≅CD as BD is a perpendicular bisector and using SAS criteria we can prove that ΔABD≅ΔCBD. BD and CD are not the corresponding sides and we do not have any other information about the sides. Hence, it is not always possible to prove that BD≅DC.